Question

Two years ago, Dilip was three times as old
as his son, and two years later, twice his age,
will be five times the age of his son..
Find their present ages​

Answer 1

Let the present age of dilip sons be x years respectively.

According to the given condition,

Two year ago

Age of Dilip son = x - 2

Age of Dilip = 3(x - 2)

Age two year

Age of Dilip son = x + 2

Age of Dilip = 3(x - 2) + 4

3x - 6 + 4

= 3x - 2

The Quetsion says that twice age of his will be equal to five times the age of his son

Therefore,

2 time the age of Dilip = 5 times the age of his son

2(3x - 2) = 5(x + 2)

6x - 4 = 5x + 10

6x - 5x = 10 + 4

x = 14

Therefore, the present age of Dilip son is 14 years.

Substituting x = 14

3(x - 2)

3(14 - 2)

3(12)

36

Therefore, the present age of Dilip is 36 years.

Answer 2

Answer :-

38 years and 14 years are present ages of Dilip and his son respectively.

Solution :-

Let the present age od Dilip be 'x' years and let the present age of his son be 'y' years

Two years ago :-

Age of Dilip = (x - 2) years

Age of his son = (y - 2) years

Given

Age of Dilip 2 years ago = 3 times the age of his son two years ago

⇒ (x - 2) = 3(y - 2)

⇒ x - 2 = 3y - 6

⇒ x - 3y = - 6 + 2

⇒ x - 3y = - 4 --eq(1)

Two years later :-

Age of Dilip = (x + 2) years

Age of his son = (y + 2) years

Given

Twice the age of Dilip two years later = Five times the age of his son two years later

⇒ 2(x + 2) = 5(y + 2)

⇒ 2x + 4 = 5y + 10

⇒ 2x - 5y = 10 - 4

⇒ 2x - 5y = 6 --eq(2)

Multiply eq(1) by 2

⇒ 2(x) - 3y(2) = - 4(2)

⇒ 2x - 6y = - 8 ---eq(3)

Subtracting eq(2) from eq(4)

⇒ 2x - 6y - (2x - 5y) = - 8 - 6

⇒ 2x - 6y - 2x + 5y = - 14

⇒ - y = - 14

⇒ y = 14

i.e Son's present age = 14 years

Substitute y = 14 in eq(1)

⇒ x - 3y = - 4

⇒ x - 3(14) = - 4

⇒ x - 42 = - 4

⇒ x = - 4 + 42

⇒ x = 38

i.e Dilip's present age = 38 years

Therefore 38 years and 14 years are present ages of Dilip and his son respectively.